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Area of Similar Triangles

Triangles II - Concepts

Calculating Perimeter of Triangle

Area of Similar Triangles

Similarity Related To Right Triangle

Pythagoras Theorem

Results Deduced From Pythagoras Theorem

Class - RRB Technician Grade I Signal Subjects

Concept Explanation

Area of Similar Triangles

Theorem: Ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

Given: $large Delta ABC sim Delta PQR$

To Prove:

$large frac{area(Delta ABC)}{area(Delta PQR)}=frac{AB^2}{PQ^2}=frac{BC^2}{QR^2}=frac{AC^2}{PR^2}$

Construction: Draw AD $large perp$ BC and PS $large perp$ QR

Proof: In $large Delta$ ABD and $large Delta$ PQS

$large angle B = angle Q$ $large [;Delta ABC sim Delta PQR;]$

$large angle D = angle S$ $large [ ;Each ;90^0;]$

$large Rightarrow Delta ABD sim Delta PQS$ [ By AA Similarity Criterion ]

Ratio of corresponding sides of similar triangle

$large frac{AB}{PQ}=frac{AD}{PS}$ ..........................(1)

But we are given that $large Delta ABC sim Delta PQR$

$large frac{AB}{PQ}=frac{AC}{PR}=frac{BC}{QR}$ ..........................(2)

From Eq (1) and (2)

$large frac{BC}{QR}=frac{AD}{PS}$

Now

$large frac{area(Delta ABC)}{area(Delta PQR)}=frac{frac{1}{2}X BC;X;AD}{frac{1}{2}X QR;X;PS}$ $large [;area ;of ;Delta = frac{1}{2};X;base;X;height;]$

$large frac{area(Delta ABC)}{area(Delta PQR)}=frac{BC;X;AD}{QR;X;PS}=frac{BC^2}{QR^2}$

As $large frac{AB}{PQ}=frac{AC}{PR}=frac{BC}{QR}$

Hence $large frac{area(Delta ABC)}{area(Delta PQR)}=frac{AB^2}{PQ^2}=frac{BC^2}{QR^2}=frac{AC^2}{PR^2}$

Illustration: If the area of two similar triangles are equal, prove that they are congruent.

Let $large Delta ABC sim Delta PQR$

$large Rightarrow large frac{AB}{PQ}=frac{AC}{PR}=frac{BC}{QR}$

Given

$large frac{area(Delta ABC)}{area(Delta PQR)}=1$

We know that

$large frac{area(Delta ABC)}{area(Delta PQR)}=frac{AB^2}{PQ^2}=frac{BC^2}{QR^2}=frac{AC^2}{PR^2}$

$large Rightarrow frac{AB^2}{PQ^2}=frac{BC^2}{QR^2}=frac{AC^2}{PR^2}=1$

$large Rightarrow$ AB = PQ , BC= QR and AC = PR

Hence

$large Delta ABC cong Delta PQR$ [By SSS Criterion of Congruence]

Sample Questions

(More Questions for each concept available in Login)

Question : 1

The ratio of any two corresponding sides of two similar triangles is 3:2; the ratio of the areas of the two triangles is _________________.
A. 4:9	B. 5:3	C. 9:4	D. 3:5
Right Option : C
View Explanation

Question : 2

The area of two isosceles triangles are in the ratio 16 : 25. The ratio of their corresponding heights is _______________
A. 5:4	B. 4:5	C. 3:2	D. 2:3
Right Option : B
View Explanation

Question : 3

In the following figure, $bigtriangleup ABCsim bigtriangleup DEF$ . The area of $bigtriangleup ABC$ is $9;cm^{2}$ and that of $bigtriangleup DEF$ is $16;cm^{2}$ , EF = 4.2 cm. Find BC.
A. 3.25 cm	B. 3.17 cm	C. 3.15 cm	D. 3.12 cm
Right Option : C
View Explanation

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Area of Similar Triangles

Area of Similar Triangles

Students / Parents Reviews [20]

Yatharthi Sharma

Prisha Gupta

Barkha Arora

Manish Kumar

Upmanyu Sharma

Aman Kumar Shrivastava

Cheshta

Barkha Arora

Aman Kumar Shrivastava

Shivam Rana

Sharandeep Singh

Aman Kumar Shrivastava

Shreya Shrivastava

Ayan Ghosh

Anika Saxena

Archit Segal

Yogansh Nyasi

Shreya Shrivastava

Manya

Aman Kumar Shrivastava